Explanation:
It is given that,
Velocity of the particle moving in straight line is :
[tex]v(t)=t^2e^{-2t}\ m/s[/tex]
We need to find the distance (x) traveled by the particle during the first t seconds. It is given by :
[tex]x=\int\limits {v.dt}[/tex]
[tex]x=\int\limits {t^2e^{-2t}dt}[/tex]
Using by parts integration, we get the value of x as :
[tex]x=\dfrac{-(2t^2+2t+1)e^{-2t}}{4}\ meters[/tex]
Hence, this is the required solution.
For a particle moving along a straight line with a velocity [tex]v(t) = t^2e^-2t[/tex] the distance traveled in meters is [tex]\frac{e^-2t}{2} [t^2 - t + 1] + C[/tex]
Let us recall that;
v = ds/dt hence;
s = [tex]\int\limits^a_b {v} \, dt[/tex]
Within the first t seconds;
s = [tex]\int\limits^t_0 {v} \, dt[/tex]
Hence, for [tex]v(t)=t^2e^-2t[/tex]
s = [tex]\int\limits^t_0{t^2e^-2t } \, dt[/tex]
Integrating by parts;
[tex]\int\limits^t_0 {u\frac{dv}{dt} } \, dt = uv - \int\limits^t_0 {v\frac{du}{dt} } \, dt[/tex]
Note that;
u = [tex]t^2[/tex]
du = [tex]2tdt[/tex]
dv = [tex]e^-2tdt[/tex]
v = [tex]\frac{e^{-2t}}{-2}[/tex]
Substituting;
[tex]\int\limits^t_0 {t^2e^{-2t} } \, dt = t^2 (e^-2t/-2) - \int\limits^t_0 {(e^-2t/-2)} \, 2tdt[/tex]
Integrating by parts, we have the result;
[tex]\frac{e^-2t}{2} [t^2 - t + 1] + C[/tex]meters
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